Question

Log

2e^-2t + 5e^-t = 3, determine the value of t.

Need someone to double check.
I got t=3, is it correct?
Thanks !!

Answer 1

can u type the equation ?

Answer 2

One way to check the solution is by putting back the value of t in equation. If solution is correct you will get zero after putting value of t.
actually e^-t =-3 or e^-t =1/2 now you can solve to get t

Answer 3

\[2e ^{-2t}+5e ^{-t}-3=0\]
\[Put e ^{-t}=x,squaring,e ^{-2t}=x ^{2}\]
\[2x ^{2}+5x+3=0\]
make factors and find the value of x i.e.,
\[e ^{-t}\]

Answer 4

you got this?

Answer 5

yea my equation is 2x^2+5x-3=0
not "+3"
am I right?
the answer I got is t= 1/3
right ?

Answer 6

not quite

Answer 7

\[ 2x^2+5x-3=0\]
\[(x+3)(2x-1)=0\] so
\[x=-3\] or \[x=\frac{1}{2}\] but you are still not done here

Answer 8

because your job was not to solve for \(x\) but for \(t\) if \(x=e^{-t}\)
this is where the log comes in

Answer 9

\[e^{-t}=-3\] has no solution as \(e\) to any power is positive

Answer 10

\[e^{-t}=\frac{1}{2}\] can you solve that?

Answer 11

1/e^t = 1/2
e^t = 2
t ln e= ln2
t = 2?

Answer 12

last line is wrong

Answer 13

\[e^{-t}=\frac{1}{2}\iff -t=\ln(\frac{1}{2})=-\ln(2)\] so
\[t=\ln(2)\]

Answer 14

or if you prefer
\[e^t=2\iff t=\ln(2)\] same thinig

Answer 15

oh so it 0.69!

Answer 16

as a decimal approximation, yes

Answer 17

thank you!!